Answer:
heat rate = 7.38 W
Explanation:
Given Data:
Pressure = 1atm
diameter (D) = 5mm = 0.005m
length = 2
mass flow rate (m) = 140*10^-6 kg/s
Exit temperature = 160°C,
At 400K,
Dynamic viscosity (μ) = 22.87 *10^-6
Prandtl number (pr) = 0.688
Thermal conductivity (k) = 33.65 *10^-3 W/m-k
Specific heat (Cp) = 1.013kj/kg.K
Step 1: Calculating Reynolds number using the formula;
Re = 4m/πDμ
= (4*140*10^-6)/(π* 0.005*22.87 *10^-6)
= 5.6*10^-4/3.59*10^-7
= 1559.
Step 2: Calculating the thermal entry length using the formula
Le = 0.05*Re*Pr*D
Substituting, we have
Le = 0.05 * 1559 * 0.688 *0.005
Le = 0.268
Step 3: Calculate the heat transfer coefficient using the formula;
Nu = hD/k
h = Nu*k/D
Since Le is less than given length, Nusselt number (Nu) for fully developed flow and uniform surface heat flux is 4.36.
h = 4.36 * 33.65 *10^-3/0.005
h = 0.1467/0.005
h = 29.34 W/m²-k
Step 4: Calculating the surface area using the formula;
A = πDl
=π * 0.005 * 2
=0.0314 m²
Step 5: Calculating the temperature Tm
For energy balance,
Qc = Qh
Therefore,
H*A(Te-Tm) = MCp(Tm - Ti)
29.34* 0.0314(160-Tm) = 140 × 10-6* 1.013*10^3 (Tm-100)
0.921(160-Tm) = 0.14182(Tm-100)
147.36 -0.921Tm = 0.14182Tm - 14.182
1.06282Tm = 161.542
Tm = 161.542/1.06282
Tm = 151.99 K
Step 6: Calculate the rate of heat transferred using the formula
Q = H*A(Te-Tm)
= 29.34* 0.0314(160-151.99)
= 7.38 W
the Prandtl number using the formula
